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Basic Chemical Calculations, Determining Chlorine Dose in Waterworks Operation Math for Water Technology MTH 082 Lecture 1 Handout on water chemistry,

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1 Basic Chemical Calculations, Determining Chlorine Dose in Waterworks Operation Math for Water Technology MTH 082 Lecture 1 Handout on water chemistry, Basic Science Chemistry Ch 3,4,7 Chapter 3 and 12- Math for Water Technology Operators Lbs/day formula, Dose Demand Residual

2 Week 2-3 Objectives 1.Review Temperature 2.Learn to calculate basic chemical solutions 3.Understand new formulas for liquid and solid chlorine application 1.Review Temperature 2.Learn to calculate basic chemical solutions 3.Understand new formulas for liquid and solid chlorine application Reading assignment: Chapter 3 and 12- Math for Water Technology Operators Chemistry Chapter 3,4 and Chapter 7 Chemistry Chemical Dosage Problems (Basic Science Concepts and Applications) Reading assignment: Chapter 3 and 12- Math for Water Technology Operators Chemistry Chapter 3,4 and Chapter 7 Chemistry Chemical Dosage Problems (Basic Science Concepts and Applications)

3 Temperature Conversions o F= (9 * o C) + 32 5 o F= (9 * o C) + 32 5 o C= 5 * ( o F – 32) 9 o C= 5 * ( o F – 32) 9 Convert 17 o C to Fahrenheit Convert 451 o F to degrees Celsius o F= (9 *17)+32=62.6 o F= 63 o F 5 o F= (9 *17)+32=62.6 o F= 63 o F 5 Celsius to Fahrenheit 1. Begin by multiplying the Celsius temperature by 9. 2. Divide the answer by 5. 3. Now add 32. Celsius to Fahrenheit 1. Begin by multiplying the Celsius temperature by 9. 2. Divide the answer by 5. 3. Now add 32. Fahrenheit to Celsius 1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 5. Fahrenheit to Celsius 1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 5. o C= 5* ( o F -32)=232.7 oC = 233 o C 9 o C= 5* ( o F -32)=232.7 oC = 233 o C 9

4 Given Formula: Solve: Given Formula: Solve: Convert 88 o F to o C? 88 o F o C= 5 * ( o F – 32) 9 o C= 5 * (88-32)/9 o C= 31 88 o F o C= 5 * ( o F – 32) 9 o C= 5 * (88-32)/9 o C= 31 o C= 5 * ( o F – 32) 9 o C= 5 * (88-32)/9 o C= 31 o C= 5 * ( o F – 32) 9 o C= 5 * (88-32)/9 o C= 31 1.31 O C 2.67 O C 3.17 O C 1.31 O C 2.67 O C 3.17 O C

5 Given Formula: Solve: Given Formula: Solve: Convert 16 o F to o C? 16 o F o C= 5 * ( o F – 32) 9 o C= 5 * (16-32)/9 o C= -9 16 o F o C= 5 * ( o F – 32) 9 o C= 5 * (16-32)/9 o C= -9 o C= 5 * ( o F – 32) 9 o C= 5 * (16-32)/9 o C= -9 o C= 5 * ( o F – 32) 9 o C= 5 * (16-32)/9 o C= -9 1.-23 O C 2.-9 O C 3.26 O C 1.-23 O C 2.-9 O C 3.26 O C

6 Given Formula: Solve: Given Formula: Solve: Convert 35 o C to o F? 35 o C o F= (9 * o C) + 32 5 oF= (9 * 35 o C) + 32 5 o F= 95 o F 35 o C o F= (9 * o C) + 32 5 oF= (9 * 35 o C) + 32 5 o F= 95 o F o F= (9 * o C) + 32 5 1.57 O F 2.51 O F 3.95 O F 4.35 O F 1.57 O F 2.51 O F 3.95 O F 4.35 O F

7 Solutions/Problems Mole (mol): chemical mass unit, defined to be 6.022 x 10 23 molecules, atoms, or some other unit. A mole of a substance is a number of grams of that substance, where the number equals the substances molecular weight. Molarity (mol/kg, molal, or M) denotes the number of moles of a given substance per liter of solvent Molarity (M)– Moles of solute Liters of solution Molarity formula Grams =(formula weight, grams/mole)(liters)(M moles/liter) M (moles/liter) = _________grams____________ (formula weight, grams/mole)(final volume, liters) Mole (mol): chemical mass unit, defined to be 6.022 x 10 23 molecules, atoms, or some other unit. A mole of a substance is a number of grams of that substance, where the number equals the substances molecular weight. Molarity (mol/kg, molal, or M) denotes the number of moles of a given substance per liter of solvent Molarity (M)– Moles of solute Liters of solution Molarity formula Grams =(formula weight, grams/mole)(liters)(M moles/liter) M (moles/liter) = _________grams____________ (formula weight, grams/mole)(final volume, liters)

8 Solutions/Problems Normality: a measure of concentration: it is equal to the number of gram equivalents of a solute per liter of solution. Depends on the valence or charge (old way) Normality formula Grams =(equival. weight, grams/equival.)(liters)(N, equivalent./liter) N (equivl./liter) = __________Grams ___________ (equival. weight, grams/equival.)(final volume, liters) Normality: a measure of concentration: it is equal to the number of gram equivalents of a solute per liter of solution. Depends on the valence or charge (old way) Normality formula Grams =(equival. weight, grams/equival.)(liters)(N, equivalent./liter) N (equivl./liter) = __________Grams ___________ (equival. weight, grams/equival.)(final volume, liters)

9 Formulas Percent Strength by Weight: Weight of solute X 100 Weight of solution Molarity formula Grams=(formula weight, grams/mole)(liters)(M moles/liter) Normality Formula Grams=(equival. weight, grams/equival.)(liters)(N, equivalent./liter) Percent Strength by Weight: Weight of solute X 100 Weight of solution Molarity formula Grams=(formula weight, grams/mole)(liters)(M moles/liter) Normality Formula Grams=(equival. weight, grams/equival.)(liters)(N, equivalent./liter)

10 _________ is defined as the number of equivalents of solute dissolved in one liter of solution. 1.Normality 2.Molarity 3.Alkalinity 4.Acidity 1.Normality 2.Molarity 3.Alkalinity 4.Acidity

11 The three most commonly used coagulants in water treatment are: 1.Aluminum hydroxide, lime and sodium hydroxide 2.Aluminum sulfate, ferric chloride, and ferrous sulfate 3.Lime, sodium hydroxide, and chlorine 4.Soda, lime and chlorine 1.Aluminum hydroxide, lime and sodium hydroxide 2.Aluminum sulfate, ferric chloride, and ferrous sulfate 3.Lime, sodium hydroxide, and chlorine 4.Soda, lime and chlorine

12 A chemical commonly used for coagulation in water treatment is: 1.Chlorine 2.Soda ash 3.Alum 4.Copper sulfate 1.Chlorine 2.Soda ash 3.Alum 4.Copper sulfate

13 The chemical symbol for the most common coagulant used in water treatment, aluminum sulfate (alum), is: 1.Al 2 (OH) 6 2.Fe 2 (SO 4 ) 3 3.NH 3 (OH) 7 4.Al 2 (SO 4 ) 3 1.Al 2 (OH) 6 2.Fe 2 (SO 4 ) 3 3.NH 3 (OH) 7 4.Al 2 (SO 4 ) 3

14 Molecular Weights Step One: Determine how many atoms of each different element are in the formula. Step Two: Look up the atomic weight of each element in a periodic table. Step Three: Multiply step one times step two for each element. Step Four: Add the results of step three together and round off as necessary. Step One: Determine how many atoms of each different element are in the formula. Step Two: Look up the atomic weight of each element in a periodic table. Step Three: Multiply step one times step two for each element. Step Four: Add the results of step three together and round off as necessary.

15 Solutions/Problems Lime calcium oxide (CaO): 1 atom of calcium= 40 grams 1 atom of oxygen= 16 grams Formula weight = 56 grams in 1 mole So a.10 mole solution would contain how many grams? (0.10)(56 grams)= 5.6 grams Lime calcium oxide (CaO): 1 atom of calcium= 40 grams 1 atom of oxygen= 16 grams Formula weight = 56 grams in 1 mole So a.10 mole solution would contain how many grams? (0.10)(56 grams)= 5.6 grams

16 Given Formula Solve: Given Formula Solve: Determine the molar mass of ALUM chemical formula Al 2 (SO 4 ) 3 ? 2 Al, 3 S, 12 O FIND Al=26.98 g, S=32.06 g, O = 16 g MM= Al 2(26.98 g) + S 3(32.06g)+ O12(16g) MM= 2Al + 3S+ 12O MM= 53.96 g + 96.18 g + 192 g MM Al 2 (SO 4 ) 3 = 342.14 g 2 Al, 3 S, 12 O FIND Al=26.98 g, S=32.06 g, O = 16 g MM= Al 2(26.98 g) + S 3(32.06g)+ O12(16g) MM= 2Al + 3S+ 12O MM= 53.96 g + 96.18 g + 192 g MM Al 2 (SO 4 ) 3 = 342.14 g 1.75 g 2.198.2 g 3.166 g 4.342.14 g 1.75 g 2.198.2 g 3.166 g 4.342.14 g

17 Given Formula Solve: Given Formula Solve: Determine the molar mass of sodium hexametaphosphate chemical formula (NaPO 3 ) 6 ? 6 NA, 6 P, 18 O FIND Na=22.98 g, P=30.97 g, O = 16 g MM= Na 6( 22.98g) + P 6(30.9g)+ O18(16g) MM= 6Na + 6P+ 18O MM= 137.88 g + 185.4 g + 288 g MM (NaPO 3 ) 6 =611.28 g 6 NA, 6 P, 18 O FIND Na=22.98 g, P=30.97 g, O = 16 g MM= Na 6( 22.98g) + P 6(30.9g)+ O18(16g) MM= 6Na + 6P+ 18O MM= 137.88 g + 185.4 g + 288 g MM (NaPO 3 ) 6 =611.28 g 1.70 g 2.611.28 g 3.700.38 g 4.50.20 g 1.70 g 2.611.28 g 3.700.38 g 4.50.20 g

18 M1V1=M2V2 M 1 V 1 = M 2 V 2 1 is starting (concentrated conditions) 2 is ending (dilute conditions) M 1 V 1 = M 2 V 2 1 is starting (concentrated conditions) 2 is ending (dilute conditions)

19 Given Formula Solve: Given Formula Solve: If we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L? M1 = 3 mol/L or 3 M, V1 = 1 L, V2 = 6 L M1V1 = M2V2, M1V1/V2 = M2 M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M M1 = 3 mol/L or 3 M, V1 = 1 L, V2 = 6 L M1V1 = M2V2, M1V1/V2 = M2 M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M V 1 = 1 L M 1 = 3 M M 1 V 1 = 3 mol V 2 = 6 L M 2 = 0.5 M M 2 V 2 = 3 mol 1.19 M 2.2 M 3.0.5 M 4.20 M 1.19 M 2.2 M 3.0.5 M 4.20 M

20 Given Formula Solve: Given Formula Solve: What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl? M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L M1V1 = M2V2, M1V1/M2 = V2 V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L M1V1 = M2V2, M1V1/M2 = V2 V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L 1.6 L 2.24 L 3.12 L 4.1 L 1.6 L 2.24 L 3.12 L 4.1 L

21 Given Formula Solve: Given Formula Solve: How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution? M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M) V1 = 0.03125 L = 31.25 mL M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M) V1 = 0.03125 L = 31.25 mL 1.437.5 mL 2.31.25 mL 3.2000 mL 4.765 mL 1.437.5 mL 2.31.25 mL 3.2000 mL 4.765 mL

22 Chlorine Concentrations 1.Sodium hypochlorite 5 to 15% available chlorine 2. Calcium hypochlorite 65-70% available chlorine 3. Chlorine gas 100% available chlorine 1.Sodium hypochlorite 5 to 15% available chlorine 2. Calcium hypochlorite 65-70% available chlorine 3. Chlorine gas 100% available chlorine

23 Determining Cl Concentrations from Hypochlorite dosage 1.Disinfection requires 280 lb/day chlorine. If CaOCl (65% available Cl) is used how many lbs day are required 1.Disinfection requires 280 lb/day chlorine. If CaOCl (65% available Cl) is used how many lbs day are required (%concentration)(X lb/day)= total lbs/day required Rearrange: (X lb /day) = (280 lbs) = 430.77 lb/d CaOCL (.65) (%concentration)(X lb/day)= total lbs/day required Rearrange: (X lb /day) = (280 lbs) = 430.77 lb/d CaOCL (.65)

24 Hypochlorite Solution Feed Rate 1. Actual Dose=Solution Feeder dose. (mg/L chlorine)(MGD)(8.34)= (mg/L chlorine)(MGD)(8.34) 1. Actual Dose=Solution Feeder dose. (mg/L chlorine)(MGD)(8.34)= (mg/L chlorine)(MGD)(8.34)

25 % Dry Strength of Solution 1. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100 Solution(lbs) 1. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100 Solution(lbs) 2. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100 Water, Lbs + Chlorine(lbs) 2. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100 Water, Lbs + Chlorine(lbs) 3. Dry chlorine % Chlorine Strength= Hypo(lbs)(% available Cl) X 100 100 Water, Lbs+ hypo(lbs) (% available Cl) 100 3. Dry chlorine % Chlorine Strength= Hypo(lbs)(% available Cl) X 100 100 Water, Lbs+ hypo(lbs) (% available Cl) 100

26 % Liquid Strength of Solution 1. Liquid chlorine Lbs Cl in liq. hypo= Lbs of chlorine in hypochlorite solution 1. Liquid chlorine Lbs Cl in liq. hypo= Lbs of chlorine in hypochlorite solution 2. Liquid chlorine (liq. Hypo lbs)(%Strength liq. Hyo) = (hypo Sol lbs) (% Strength liq. hyp) 100 100 2. Liquid chlorine (liq. Hypo lbs)(%Strength liq. Hyo) = (hypo Sol lbs) (% Strength liq. hyp) 100 100 3. Liquid chlorine (liq. Hypo gal)(8.34)(%Strength liq. Hyo)=(hypo Sol gal)(8.34)(%Strength liq. Hyp) 100100 3. Liquid chlorine (liq. Hypo gal)(8.34)(%Strength liq. Hyo)=(hypo Sol gal)(8.34)(%Strength liq. Hyp) 100100

27 Given Formula Solve: Given Formula Solve: A chlorinator setting of 20 lbs of chlorine per 24 hrs results in a residual of 0.4 mg/L. The chlorinator setting is 25 lb per 24 hrs. The chlorine residual increased to 0.5 mg/L at this new dosage rate. The average flow being treated is 1.6 mgd. On the basis of this data is the water being chlorinated beyond breakpoint? Residual 1= 0.4 mg/L, residual 2=0.5 mg/L, 1.6 mgd, new 5 lb/day Lbs/d incr= Dose( flow)(8.34 lb/g) Act increase in residual=New residual-Old residual Dose= Lb/day increase/flow 8.34 lb/g 5 lb/day/1.6 mgd(8.34 lb/d)= Residual = 0.37 mg/L Actual increase in residual was 0.5 mg/L -0.4 mg/L =0.1 mg/L Expected was 0.37 but the actual was 0.1. Not being met! Residual 1= 0.4 mg/L, residual 2=0.5 mg/L, 1.6 mgd, new 5 lb/day Lbs/d incr= Dose( flow)(8.34 lb/g) Act increase in residual=New residual-Old residual Dose= Lb/day increase/flow 8.34 lb/g 5 lb/day/1.6 mgd(8.34 lb/d)= Residual = 0.37 mg/L Actual increase in residual was 0.5 mg/L -0.4 mg/L =0.1 mg/L Expected was 0.37 but the actual was 0.1. Not being met! 1.yes 2.no 1.yes 2.no

28 Specific Gravity, LBS, Gallons, Solution Strength

29 Team Scores

30 Today’s objective: Review basic chemistry and solution making as it pertains to the waterworks industry Calculate the chemical dosage using the standard “pounds formula” has been met? 1.Strongly Agree 2.Agree 3.Disagree 4.Strongly Disagree 1.Strongly Agree 2.Agree 3.Disagree 4.Strongly Disagree

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